Question

An object is moving with a velocity of * 6 ms and
with an acceleration of -1 ms 2. What will be the
(a) distance travelled by the car before coming to
rest?
(b) time taken for coming to rest?​

Answer 1

Answer:

a) Distance travelled is 18 m

b) Time taken is 6 seconds.

Explanation:

Given:

Velocity of the object = 6 m/s²

Acceleration = –1 m/s²

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• Question a)

★ Using the equation of motion :

$\rightarrow$ 2as = v² – u²

• v = 0 m/s (as the object will come to rest)
• u = 6 m/s
• a = –1 m/s²
• s = distance

$\rightarrow$ 2(–1)s = (0)² – (6)²

$\rightarrow$ –2s = 0 – 36

$\rightarrow$ –2s = –36

$\rightarrow$ s = –36/–2

$\rightarrow$ s = 18 m

Distance travelled by the car before coming to rest is 18 m.

___________________

• Question b)

★ Using the equation of motion :

$\rightarrow$ v = u + at

• v = 0 m/s
• u = 6 m/s
• a = –1 m/s²
• t = ?

$\rightarrow$ v = u + at

$\rightarrow$ 0 = 6 + (-1) × t

$\rightarrow$ 0 = 6 - t

$\rightarrow$ t = 6 seconds

Time taken to come to rest is 6 seconds.

Answer 2

Given : Initial Velocity ( u ) of the object is 6 m/s & Acceleration ( a ) is -1 m/s² .

Exigency To Find : (a) Distance travelled by the car before coming to rest ? & (b) Time taken for coming to rest ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Answer of a : Distance travelled by the car before coming to rest ?

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀▪︎⠀⠀Initial Velocity ( u ) of the object is 6 m/s

⠀⠀⠀⠀▪︎⠀⠀Acceleration ( a ) of object is -2 m/s²

⠀⠀⠀⠀▪︎⠀ ⠀Final Velocity ( v ) of the object is 0 m/s [ as , we have to find before rest ]

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\\qquad\maltese\:\:\bf Third \:Equation\:of \:motion \::$

$\qquad \dag\:\:\bigg\lgroup \pmb{ \bf v^2 = u^2 + 2as }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , v is the Final velocity , u is the Initial velocity a is the Acceleration & s is the Distance Travelled

$\qquad \dashrightarrow \sf v^2 = u^2 + 2as \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf v^2 = u^2 + 2as \:$

$\qquad \dashrightarrow \sf (0)^2 = (6)^2 + 2(-1)s \:$

$\qquad \dashrightarrow \sf 0 = (6)^2 + 2(-1)s \:$

$\qquad \dashrightarrow \sf 0 = 36 + 2(-1)s \:$

$\qquad \dashrightarrow \sf 0 = 36 + (-2)s \:$

$\qquad \dashrightarrow \sf 0 - 36 = (-2)s \:$

$\qquad \dashrightarrow \sf - 36 = -2s \:$

$\qquad \dashrightarrow \sf s = \:\dfrac{-36}{-2}\:$

$\qquad \dashrightarrow \sf s = \:\dfrac{\cancel{-}36}{\cancel{-}2}\:$

$\qquad \dashrightarrow \sf s = \:\cancel {\dfrac{36}{2}}\:$

$\qquad \dashrightarrow \sf s = \:18\:$

$\qquad \therefore \pmb{\underline{\purple{\frak{ \:s\: = \:18 \: m \: }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀⠀▪︎⠀Here , s denotes Distance Travelled by car before coming to rest which is 18 m .

$\therefore \:\underline { \sf Hence , \:\: The \:Distance \:Travelled \:is \: \bf 18 \: m \:}.$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Answer of b : Time taken for coming to rest ?

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀▪︎⠀⠀Initial Velocity ( u ) of the object is 6 m/s

⠀⠀⠀⠀▪︎⠀⠀Acceleration ( a ) of object is -2 m/s²

⠀⠀⠀⠀▪︎⠀ ⠀Final Velocity ( v ) of the object is 0 m/s [ as , we have to find before rest ]

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\\qquad\maltese\:\:\bf First \:Equation\:of \:motion \::$

$\qquad \dag\:\:\bigg\lgroup \pmb{ \bf v = u + at }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , v is the Final velocity , u is the Initial velocity a is the Acceleration & t is the Time taken

$\qquad \dashrightarrow \sf v = u + at \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf v = u + at \:$

$\qquad \dashrightarrow \sf 0 = 6 + (-1)t \:$

$\qquad \dashrightarrow \sf 0 = 6 - t \:$

$\qquad \dashrightarrow \sf t = 6 \:$

$\qquad \therefore \pmb{\underline{\purple{\frak{ \:t\: = \:6 \: seconds\: }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀⠀▪︎ Here , t is the time taken before coming rest which is 6 Seconds

$\therefore \:\underline { \sf Hence , \:\: The \:Time \:Taken \:is \: \bf 6 \: seconds \:}.$